01.03 Module One Quiz

1.) In Splotts 4th row he made a mistake instead of adding 4 to both sides you would subtract 4 from both sides making the answer 5x=10 not 5x=18.Then you divide 5 from both sides and end up with x=2. In the 5th column of fizzles work he puts -12=-5x but, he subtracted wrong the answer should be -10=-5x because 4-14 equals -10 not -12, And after that you wouldn’t add 5 to both sides you would divide 5 from both sides leaving x to equal 2.

2.) F(x)=2x+4

3.) My function is legitimate because you can plug x or y into it and it is in function form.

4.) To solve for f(3) you must plug in 3 for x. So using my function F(x)=2x+4 I need to replace my x’s so my new equation will be f(3)=2(3)+4. Than just multiply 2 and 3 which gives you 6 and add 4 so, f(3)=10.

5.) To find the inverse of my function you must put y in to replace f(x). So y=2x+4 then, switch the x and y places x=2y+4. After that, subtract 4 from both sides x-4=2y+4-4. You end up with x-4=2y. Now divide 2 from both sides you will have 2 under x-4 = to just y because the 2’s cancel out. Now you must replace y with f –1(x). Replacing the x’s with a 3. Your equation should be 3-4 over 2 equaled to f –1(3). 3-4=-1. -1 over 2 equals -0.5.

6.) My new function would be h(x) =-3x+14 and I will assign 2 in for x. So, h(2)=-3(2)+14 would equal h(2)=8. Using the formula f(h(2)) now I plug 8 into f(x). F(8)=2(8)+4=16+4=20. Now repeating the process I recall that f(3)=10 so, im going to plug 10 into x for h(x). I get h(10)=-3(10)+14=-30+14=-16. So when you compare f(h(2)) and h(f(2)) the first one equals 20 and the second equals -16 proving these numbers are not equal and that f(h(x)) and h(f(x)) are not always the same number.